. Ionic Product of Water (Edexcel International A Level Chemistry)

Revision Note

Sonny

Author

Sonny

Last updated

Ionic Product of Water, Kw

  • In all aqueous solutions, an equilibrium exists in water where a few water molecules dissociate into protons and hydroxide ions
  • We can derive an equilibrium constant for the reaction:

Deriving Kw, downloadable AS & A Level Chemistry revision notes

  • This is a specific equilibrium constant called the ionic product for water
  • The product of the two ion concentrations is always 1 x 10-14 moldm-6
  • This makes it straightforward to see the relationship between the two concentrations and the nature of the solution:

[H+] & [OH] Table

[H+] and [OH-] table, downloadable IB Chemistry revision notes

The relationship between Kand pKw is given by the following equation:

pKw = -logKw

pH Calculation of a Strong Base

  • Strong bases are completely ionised in solution

BOH (aq) → B+ (aq) + OH- (aq)

  • Therefore, the concentration of hydroxide ions [OH-] is equal to the concentration of base [BOH]
    • Even strong alkalis have small amounts of H+ in solution which is due to the ionisation of water
  • The concentration of OH- in solution can be used to calculate the pH using the ionic product of water
  • Once the [H+] has been determined, the pH of the strong alkali can be founding using pH = -log[H+]

Finding pH of strong bases, downloadable AS & A Level Chemistry revision notes

  • Similarly, the ionic product of water can be used to find the concentration of OH- ions in solution if [H+] is known, simply by dividing Kw by the [H+]

Worked example

pH calculations of a strong alkali

Question 1: 

Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH

Question 2: 

Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50

Answer

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na+ (aq) + OH- (aq) 

Answer 1:

   The pH of the solution is:

   [H+] = K ÷ [OH-]

   [H+] = (1 x 10-14) ÷ 0.15 = 6.66 x 10-14

   pH = -log[H+]

         = -log 6.66 x 10-14  = 13.17

Answer 2

Step 1: Calculate hydrogen concentration by rearranging the equation for pH

   pH = -log[H+]

   [H+]= 10-pH

   [H+]= 10-10.50

   [H+]= 3.16 x 10-11 mol dm-3

Step 2: Rearrange the ionic product of water  to find the concentration of hydroxide ions

   Kw = [H+] [OH-]

    [OH-]K ÷  [H+

Step 3: Substitute the values into the expression to find the concentration of hydroxide ions

   Since Kw is 1 x 10-14 mol2 dm-6,

    [OH-]= (1 x 10-14)  ÷  (3.16 x 10-11)

   [OH-]= 3.16 x 10-4 mol dm-3

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Sonny

Author: Sonny

Expertise: Chemistry

Sonny graduated from Imperial College London with a first-class degree in Biomedical Engineering. Turning from engineering to education, he has now been a science tutor working in the UK for several years. Sonny enjoys sharing his passion for science and producing engaging educational materials that help students reach their goals.